Question

Explain formation of \([CoF_6]^{3-}\) complex with respect to
i. Hybridisation ii. Magnetic properties
iii. Inner/outer complex iv. Geometry 
 

Hint:

In coordination complexes, the geometry and magnetic properties depend on the metal ion's oxidation state, the ligands, and their arrangement. Understanding the hybridization and electron configuration helps determine these properties.

Answer 1

i. Hybridisation:
In the complex \([CoF_6]^{3-}\), cobalt is in the +3 oxidation state (Co$^{3+}$), which has an electronic configuration of [Ar] \(3d^6\). The \(Co^{3+}\) ion undergoes \(sp^3d^2\) hybridization to form six bonds with fluoride ions. ii. Magnetic Properties:
The \([CoF_6]^{3-}\) complex has no unpaired electrons because all the electrons are paired in the \(3d\) orbitals. Therefore, the complex is diamagnetic. iii. Inner/Outer Complex:
The fluoride ions in \([CoF_6]^{3-}\) are considered ligands that directly coordinate with the cobalt ion, so this is an inner complex. iv. Geometry:
The geometry of the \([CoF_6]^{3-}\) complex is octahedral, as the six fluoride ions are arranged symmetrically around the central cobalt ion.